Question

for each part below, solve the equation. (a) solve for x. 3(x + 1)+2x = 5(x - 1)+5 no solution x = all real numbers are solutions (b) solve for v. -4(v + 2)+2 = 2(v + 6) no solution v = all real numbers are solutions

Explanation:

Step1: Expand both sides of equation (a)

\[

$$\begin{align*} 3(x + 1)+2x&=5(x - 1)+5\\ 3x+3 + 2x&=5x-5 + 5\\ 5x+3&=5x \end{align*}$$

\]

Step2: Subtract \(5x\) from both sides

\[

$$\begin{align*} 5x+3-5x&=5x-5x\\ 3&=0 \end{align*}$$

\]
Since \(3
eq0\), there is no solution for \(x\).

Step3: Expand both sides of equation (b)

\[

$$\begin{align*} -4(v + 2)+2&=2(v + 6)\\ -4v-8 + 2&=2v+12\\ -4v-6&=2v+12 \end{align*}$$

\]

Step4: Add \(4v\) to both sides

\[

$$\begin{align*} -4v-6+4v&=2v+12+4v\\ -6&=6v + 12 \end{align*}$$

\]

Step5: Subtract 12 from both sides

\[

$$\begin{align*} -6-12&=6v+12-12\\ -18&=6v \end{align*}$$

\]

Step6: Divide both sides by 6

\[

$$\begin{align*} \frac{-18}{6}&=\frac{6v}{6}\\ v&=- 3 \end{align*}$$

\]

Answer:

(a) No solution
(b) \(v=-3\)